NEET-XII-Physics
38: Electromagnetic Induction
- #31A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced (a) in the loop abc, (b) in the segment bc, (c) in the segment ac and (d) in the segment ab.
FigureAns : (a) The emf induced in loop abc is zero, as there is no change in the magnetic flux through it. (b) The emf induced is given by
`` e=\left(\stackrel{\to }{v}\times \stackrel{\to }{B}\right).\stackrel{\to }{l}``
Emf induced in segment bc, e = Bvl (With positive polarity at point C) (c) There is no emf induced in segment bc, as the velocity is parallel to its length. (d) The emf induced in segment ab is calculated by the following formula:
e = B.v. (Effective length of ab)
The effective length of ab is along the direction perpendicular to its velocity.
Emf induced, e = B.v.(bc)
Page No 308:
- #31-ain the loop abc,Ans : The emf induced in loop abc is zero, as there is no change in the magnetic flux through it.
- #31-bin the segment bc,Ans : The emf induced is given by
`` e=\left(\stackrel{\to }{v}\times \stackrel{\to }{B}\right).\stackrel{\to }{l}``
Emf induced in segment bc, e = Bvl (With positive polarity at point C)
- #31-cin the segment ac andAns : There is no emf induced in segment bc, as the velocity is parallel to its length.
- #31-din the segment ab.
FigureAns : The emf induced in segment ab is calculated by the following formula:
e = B.v. (Effective length of ab)
The effective length of ab is along the direction perpendicular to its velocity.
Emf induced, e = B.v.(bc)
Page No 308: